Default lambda as templated parameter of a function

Question

Consider the following code

template void foo(const T& t = []() {}) {
  // implementation here
}

void bar() {
  foo([&         


        

Answer 1

Another (very efficient) way - default T to be a null functor.

// no_op is a function object which does nothing, regardless of how many
// arguments you give it. It will be elided completely unless you compile with
// -O0
struct no_op 
{ 
    template
    constexpr void operator()(Args&&...) const {} 
};

// foo defaults to using a default-constructed no_op as its function object
template void foo(T&& t = T()) 
{    
  // implementation here
    t();
}

void bar() {
  foo([&](){ std::cout << "something\n"; }); // this compiles
  foo(); // this now compiles
}